Page 251 - Linear Algebra Done Right

Chapter 10. Trace and Determinant
242
contains D k σ j (x) in row j, column k (we will not prove this). In other
words,
 
D 1 σ 1 (x) . . . D n σ 1 (x)
 . . 
10.39 M(σ (x)) =  . . . .  .

 
D 1 σ n (x) . . . D n σ n (x)
Suppose that σ is differentiable at each point of Ω and that σ is
injective on Ω. Let f be a real-valued function defined on σ(Ω). Let
x ∈ Ω and let Γ be a small subset of Ω containing x. As we noted above,

volume σ(Γ) ≈ volume σ (x) (Γ),
where the symbol ≈ means “approximately equal to”. Using 10.38, this
becomes
volume σ(Γ) ≈|det σ (x)|(volume Γ).

Let y = σ(x). Multiply the left side of the equation above by f(y) and

the right side by f σ(x) (because y = σ(x), these two quantities are
equal), getting

10.40 f(y) volume σ(Γ) ≈ f σ(x) |det σ (x)|(volume Γ).

Now divide Ω into many small pieces and add the corresponding ver-
sions of 10.40, getting

10.41 f(y) dy = f σ(x) |det σ (x)| dx.

σ(Ω) Ω
This formula was our goal. It is called a change of variables formula
because you can think of y = σ(x) as a change of variables.
The key point when making a change of variables is that the factor
of |det σ (x)| must be included, as in the right side of 10.41. We finish

up by illustrating this point with two important examples. When n = 2,
If you are not familiar we can use the change of variables induced by polar coordinates. In this
with polar and case σ is defined by
spherical coordinates,
skip the remainder of σ(r, θ) = (r cos θ, r sin θ),
this section.
where we have used r, θ as the coordinates instead of x 1 ,x 2 for reasons
that will be obvious to everyone familiar with polar coordinates (and
will be a mystery to everyone else). For this choice of σ, the matrix of
partial derivatives corresponding to 10.39 is